Homotopy between composites associated in different ways: Difference between revisions

Revision as of 18:57, 18 December 2010

Statement

Existential version

Suppose $f_{1}, f_{2}, f_{3}$ are loops based at a point $x_{0}$ in a topological space $X$ . We can consider two differently associated products of these three loops:

$a = (f_{1} * f_{2}) * f_{3}, b = f_{1} * (f_{2} * f_{3})$

$a$ and $b$ are homotopic loops, i.e., they are in the same homotopy class of loops based at $x_{0}$ .

Constructive/explicit version

We first note the explicit piecewise definitions of $a$ and $b$ :

$a = {\begin{matrix} f_{1} (4 t), & 0 \leq t \leq 1 / 4 \\ f_{2} (4 t - 1), & 1 / 4 < t \leq 1 / 2 \\ f_{3} (2 t - 1), & 1 / 2 < t \leq 1 \end{matrix}$

and:

$b = {\begin{matrix} f_{1} (2 t), & 0 \leq t \leq 1 / 2 \\ f_{2} (4 t - 2), & 1 / 2 < t \leq 3 / 4 \\ f_{3} (4 t - 3), & 3 / 4 < t \leq 1 \end{matrix}$

If we denote the homotopy by $H$ , we want $H (t, 0) = a (t), H (t, 1) = b (t)$ and $H (0, s) = H (1, s) = x_{0}$ . This homotopy is explicitly given by Fill this in later

Graphical version

Fill this in later

@@ Line 10: / Line 10: @@
 ===Constructive/explicit version===
+We first note the explicit piecewise definitions of <math>a</math> and <math>b</math>:
+<math>a = \lbrace\begin{array}{rl} f_1(4t), & 0 \le t \le 1/4 \\ f_2(4t - 1), & 1/4 < t \le 1/2 \\ f_3(2t - 1), & 1/2 < t \le 1 \\\end{array}</math>
+and:
+<math>b = \lbrace\begin{array}{rl} f_1(2t), & 0 \le t \le 1/2 \\ f_2(4t - 2), & 1/2 < t \le 3/4 \\ f_3(4t - 3), & 3/4 < t \le 1 \\\end{array}</math>
+If we denote the homotopy by <math>H</math>, we want <math>H(t,0) = a(t), H(t,1) = b(t)</math> and <math>H(0,s) = H(1,s) = x_0</math>. This homotopy is explicitly given by {{fillin}}
+===Graphical version===
 {{fillin}}